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SRMJEEE Chemistry Mock Test - 6 - JEE MCQ


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30 Questions MCQ Test - SRMJEEE Chemistry Mock Test - 6

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SRMJEEE Chemistry Mock Test - 6 - Question 1

Which of the following will not from a yellow precipitate on heating with an alkaline solution of iodine?

SRMJEEE Chemistry Mock Test - 6 - Question 2

Schiff's reagent is

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 2

Schiff's reagent is a chemical solution used primarily in organic chemistry for detecting aldehydes. Its composition includes:

  • Rosaniline hydrochloride - a dye that forms the basis of Schiff's reagent.
  • It is often used in a test known as the Schiff test to identify the presence of aldehydes in various substances.
  • The reagent changes colour, typically to a magenta hue, when it reacts with aldehydes.
SRMJEEE Chemistry Mock Test - 6 - Question 3

Ascorbic acid is a chemical name of

SRMJEEE Chemistry Mock Test - 6 - Question 4

The IUPAC name of (C2H5)4C is

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 4

The IUPAC name of (C2H5)4C is:

  • The correct name is Tetraethylmethane.
  • This compound consists of four ethyl groups attached to a central carbon atom.
  • It is important to identify the structure correctly to assign the proper name.
SRMJEEE Chemistry Mock Test - 6 - Question 5

When H₂S gas is passed through the HCl containing aqueous solution of CuCl₂, HgCl₂, BiCl₃ and coCl₂, it does not precipitate out

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 5

When H₂S gas is passed through HCl containing aqueous solutions of CuCl₂, HgCl₂, BiCl₃, and CoCl₂, no precipitation occurs.

  • CuS: Does not form because Cu²⁺ ions are soluble in HCl.
  • HgS: Remains soluble in the presence of hydrochloric acid.
  • Bi₂S₃: Does not precipitate due to the solubility of Bi³⁺ ions in the acidic solution.
  • CoS: Co²⁺ ions are also soluble in HCl, preventing precipitation.

In summary, the presence of hydrochloric acid keeps these metal ions in solution, preventing the formation of their corresponding sulphides.

SRMJEEE Chemistry Mock Test - 6 - Question 6

Heaviest particle is

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 6

The heaviest particle among the options provided is the neutron.

  • The neutron is a subatomic particle found in the nucleus of an atom.
  • It has a mass of about 1.675 × 10-27 kg, making it heavier than both the proton and the electron.
  • Mesons are also subatomic particles, but they are generally lighter than neutrons.
  • The electron, being a fundamental particle, is significantly lighter than both neutrons and protons.
SRMJEEE Chemistry Mock Test - 6 - Question 7

Electrophile in the case of chlorination of benzene in presence of FeCl₃ is

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 7

The electrophile involved in the chlorination of benzene in the presence of FeCl₃ is the positively charged chlorine ion.

  • The reaction begins with the formation of the electrophile.
  • In this case, FeCl₃ interacts with chlorine (Cl₂) to generate Cl⁺.
  • This Cl⁺ ion is the active species that reacts with benzene.
  • The benzene ring undergoes electrophilic substitution, resulting in chlorobenzene.
SRMJEEE Chemistry Mock Test - 6 - Question 8

The greater the s- character in an orbital the ... is its energy

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 8

The greater the s- character in an orbital, the shorter is the bond and greater is its strength. Thus, bond formed by sp hybridisation , as in acetylene is stronger as compare to ethane and ethene , because it has maximum bond energy.

SRMJEEE Chemistry Mock Test - 6 - Question 9

The acid showing salt like character in aqueous solutions is

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 9

The acid that exhibits salt-like characteristics in aqueous solutions is often referred to as having a unique behaviour due to its ability to form salts when reacting with bases.

  • Acetic acid is a weak acid commonly found in vinegar.

  • Benzoic acid is known for its preservative properties in food.

  • Formic acid is often used in leather production and as a preservative.

  • α-aminoacetic acid (also known as glycine) is the simplest amino acid.

Among these options, the one that typically shows salt-like behaviour in water is most notable due to its ability to dissociate and form ionic species.

SRMJEEE Chemistry Mock Test - 6 - Question 10

Which of the following is aromatic?

SRMJEEE Chemistry Mock Test - 6 - Question 11

Sulphuric acid provides a simple example of

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 11

Sulphuric acid serves as a straightforward example of a chemical compound that exhibits specific bonding characteristics. It is primarily known for:

  • Covalent bonding between sulphur and oxygen atoms.
  • Its ability to form ionic species when dissolved in water.
  • Reactivity with various substances, highlighting its acidic nature.

Understanding sulphuric acid's properties is essential for grasping fundamental concepts in chemistry.

SRMJEEE Chemistry Mock Test - 6 - Question 12

60 J of heat flows out from 600 g of water at 30°C into surroundings at 25°C.The net entropy change in universe is approximately

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 12

To calculate the net entropy change in the universe, consider the following steps:

  • Heat loss from water: 60 J flows from 600 g of water at 30°C.
  • Temperature of water: 30°C converts to Kelvin as 303 K.
  • Change in entropy of water:
    • Using the formula: ΔS = Q/T.
    • ΔS_water = -60 J / 303 K = -0.198 J/K.
  • Temperature of surroundings: 25°C converts to Kelvin as 298 K.
  • Change in entropy of surroundings:
    • Using the formula: ΔS = Q/T.
    • ΔS_surroundings = 60 J / 298 K = 0.201 J/K.
  • Net entropy change in the universe:
    • ΔS_universe = ΔS_water + ΔS_surroundings.
    • ΔS_universe = -0.198 J/K + 0.201 J/K = 0.003 J/K.

The net entropy change in the universe is approximately 0.003 J/K, indicating a slight increase in overall entropy.

SRMJEEE Chemistry Mock Test - 6 - Question 13

A catalyst is used to

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 13

A catalyst is a substance that speeds up a chemical reaction without being consumed in the process. Its main functions include:

  • Increasing the rate of reactions, allowing products to form more quickly.
  • It does not affect the equilibrium of the reaction or change the final amounts of reactants and products.
  • Catalysts are essential in many industrial processes to improve efficiency and reduce energy costs.
SRMJEEE Chemistry Mock Test - 6 - Question 14

A glass bulb is filled with NO₂ gas and immersed in an ice bath at 0°C which becomes colourless after some time . This colourless gas will be

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 14

When a glass bulb containing NO₂ gas is placed in an ice bath at 0°C, it eventually becomes colourless. This transformation indicates the presence of a different gas.

  • Initially, NO₂ is a brown gas.
  • At low temperatures, NO₂ can convert to N₂O₄, which is colourless.
  • This process is a result of the equilibrium between NO₂ and N₂O₄.
  • Thus, the colourless gas observed is N₂O₄.
SRMJEEE Chemistry Mock Test - 6 - Question 15

Calculate the heat of formation Δ H of CO (in kcal) from the following data :

C(graphite) + O2(g) → CO2 (g); ΔH = -94K. cal
CO(g) +  (g) → CO2 (g) ; ΔH = -68K. cal

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 15

To calculate the heat of formation (ΔH) of CO:

We will use the provided thermochemical equations:

  • C(graphite) + O2(g) → CO2(g); ΔH = -94 kcal
  • CO(g) + 1/2 O2(g) → CO2(g); ΔH = -68 kcal

To find ΔH for the formation of CO, we can rearrange the equations:

  • From the first equation, we can derive the formation of CO by reversing the second equation:
  • ΔH for CO = ΔH for CO2 - ΔH for C + O2

Substituting in the values:

  • ΔH for CO = (-68 kcal) - (-94 kcal)
  • ΔH for CO = -68 + 94 kcal
  • ΔH for CO = 26 kcal

Thus, the heat of formation of CO is -26 kcal.

SRMJEEE Chemistry Mock Test - 6 - Question 16

Aspirin is an acetylation product of

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 16

Aspirin is formed from the acetylation of a specific compound. The correct source compound is:

  • o-hydroxybenzoic acid, also known as salicylic acid, is the precursor to aspirin.
  • Aspirin is created by adding an acetyl group to the hydroxyl group of salicylic acid.
  • This chemical modification enhances its properties and reduces side effects.
SRMJEEE Chemistry Mock Test - 6 - Question 17

The IUPAC name of

is

SRMJEEE Chemistry Mock Test - 6 - Question 18

The lightest metal is

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 18

Lithium is lightest among all the metals with density 0.534 g/cm3.

SRMJEEE Chemistry Mock Test - 6 - Question 19

Which one of the following reactions occur at the cathode?

SRMJEEE Chemistry Mock Test - 6 - Question 20

One mole of argon gas will have least density at

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 20

One mole of argon gas will have the least density at:

  • Density of a gas is affected by temperature and pressure.
  • Generally, gas density decreases with an increase in temperature and decreases with an increase in pressure.
  • At standard temperature and pressure (STP), argon has a certain density.
  • In conditions of higher temperature (like 273°C) at 1 atm or 2 atm, the density will be lower than at STP.
  • Thus, the conditions that yield the lowest density for argon gas will be at the highest temperature and the lowest pressure.
SRMJEEE Chemistry Mock Test - 6 - Question 21

Which of the following applies to the reaction,

1. CH₃CH = CHCH₃(major product)
2. CH₂ = CH − CH₂CH₃(minor product)

SRMJEEE Chemistry Mock Test - 6 - Question 22

A compound having the formula NH₂CH₂COOH may behave

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 22

The compound with the formula NH₂CH₂COOH can exhibit both acidic and basic properties due to its functional groups. Here are the key points:

  • The presence of the carboxylic acid group (-COOH) allows it to donate a proton (H⁺), making it act as an acid.
  • The amino group (-NH₂) has the ability to accept a proton, enabling it to function as a base.
  • As a result, this compound is classified as an amphoteric substance, capable of reacting in both acidic and basic contexts.
SRMJEEE Chemistry Mock Test - 6 - Question 23

A symmetrical molecule has 2 chiral carbon atoms. The no. of isomers is

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 23
Explanation: For a molecule to be chiral, it must have a stereocenter and no axis of symmetry. An atom with a stereocenter has no identical bonds; it is a carbon atom with four unique substituents. There are two stereocenters in each of the three molecules.
SRMJEEE Chemistry Mock Test - 6 - Question 24

C-14 has a half-life of 5760 years. 100 mg of a sample containing C-14 is reduced to 25 mg in

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 24

C-14 Half-Life Calculation

To determine how long it takes for 100 mg of C-14 to reduce to 25 mg, we can use the concept of half-lives.

  • The half-life of C-14 is 5760 years.
  • Starting with 100 mg, the sample reduces as follows:
    • After the first half-life: 100 mg → 50 mg
    • After the second half-life: 50 mg → 25 mg
  • Since it takes two half-lives to reach 25 mg, we calculate the total time:
    • 2 half-lives × 5760 years = 11520 years

Thus, it takes 11520 years for the sample to reduce from 100 mg to 25 mg.

SRMJEEE Chemistry Mock Test - 6 - Question 25

Lacrymotor or tear gas is

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 25

Lacrymotor or tear gas is

  • The chemical compound responsible for producing tears is often used for crowd control.
  • It can cause temporary blindness, difficulty breathing, and skin irritation.
  • Commonly, it is a type of chemical agent that disrupts normal functioning.
  • Safety precautions should be taken when handling or being exposed to it.
SRMJEEE Chemistry Mock Test - 6 - Question 26

Aniline reacts with concentrated HNO₃ to give :

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 26

Aniline reacts with concentrated HNO₃ to produce:

  • Nitroaniline: The primary product formed is nitroaniline, where a nitro group replaces one of the hydrogen atoms on the aniline ring.
  • The reaction proceeds via electrophilic aromatic substitution, where aniline acts as a nucleophile.
  • Concentrated nitric acid is a strong oxidising agent, promoting the formation of the nitro group.

Different isomers of nitroaniline can be formed depending on the position of the nitro group on the benzene ring:

  • Ortho-nitroaniline: Nitro group at the 2-position.
  • Meta-nitroaniline: Nitro group at the 3-position.
  • Para-nitroaniline: Nitro group at the 4-position.

The major product in this reaction is typically para-nitroaniline due to steric factors and stability of the product. The reaction conditions can influence the distribution of the isomers formed.

SRMJEEE Chemistry Mock Test - 6 - Question 27

Which is incorrect representation?

SRMJEEE Chemistry Mock Test - 6 - Question 28

Which of the following is used for making horses, shoe heels and stoppers?

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 28

The material used for making horses, shoe heels, and stoppers is:

  • Buna-S is the most suitable choice for these applications.
  • It is a type of synthetic rubber known for its durability and flexibility.
  • Other options like PMMA, styron, and neoprene are less relevant for these specific uses.
SRMJEEE Chemistry Mock Test - 6 - Question 29

In Duma's method, the gas which is collected in Nitrometer is

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 29

Duma's method collects gas in a Nitrometer, specifically aiming to identify the composition of nitrogen in a sample. The key gas collected in this process is:

  • N₂ (Nitrogen gas) - This gas is primarily collected as it is a major component of the atmosphere.
  • Other gases such as NO, NH₃, and H₂ can be involved in chemical reactions but are not the main focus of Duma's method.

The primary aim of this method is to accurately measure the nitrogen content in organic compounds.

SRMJEEE Chemistry Mock Test - 6 - Question 30

The oxidation number of Pt in [Pt(C₂H₄)Cl₃]- is

Detailed Solution for SRMJEEE Chemistry Mock Test - 6 - Question 30

The oxidation number of Pt in [Pt(C₂H₄)Cl₃]- can be determined by following these steps:

  • The overall charge of the complex is -1.
  • Each chloride ion (Cl) has an oxidation number of -1.
  • With three Cl ions, the total contribution is -3 from chlorides.
  • Ethylene (C₂H₄) is a neutral ligand, contributing 0.
  • Let the oxidation number of Pt be x.
  • The equation becomes: x + (-3) + 0 = -1.
  • Solving for x gives: x = +2.

Thus, the oxidation number of Pt in this complex is +2.

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